Question

In a marketing survey, a random sample of 998 supermarket shoppers revealed that 276 always stock up on an item when they find that item at a real bargain price.

(a) Let p represent the proportion of all supermarket shoppers who always stock up on an item when they find a real bargain. Find a point estimate for p. (Use 3 decimal places.)


(b) Find a 95% confidence interval for p. (Use 3 decimal places.)

lower limit
upper limit

(c) What is the margin of error based on a 95% confidence interval? (Use 3 decimal places.)

Answer 1

Solution :

Given that n = 998, x = 276

(a)
=> proportion p = x/n

= 276/998

= 0.2766

= 0.277 (rounded)

(b)
=> p = 0.277 ; q = 1 - p = 0.723

=> for 95% confidence interval, Z = 1.96

=> A 95% confidence interval of the proportion p is

=> p +/- Z*sqrt(p*q/n)

=> 0.277 +/- 1.96*sqrt(0.277*0.723/998)

=> 0.2492 < p < 0.3048

=> 0.249 < p < 0.305 (rounded)

=> Lower limit = 0.249

=> Upper limit = 0.305

(c)
=> Margin of error E = Z*sqrt(p*q/n)

= 1.96*sqrt(0.277*0.723/998)

= 0.0278

= 0.028 (rounded)